Download Lecture 7.1

Welcome back to Physics 215
Today’s agenda:
• Circular motion
• Impulse and momentum
Physics 215 – Fall 2014
Lecture 07-1
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Current homework assignment
• HW6:
– Knight textbook Ch.8: 36, 46, 48;
– Ch.9: 28, 48, 70
– due Friday, Oct. 10th in recitation
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Forces in circular motion
• Motion around circular track, constant speed (for now):
arad = v2/r
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Two identical balls are connected by a string and whirled
around in circles of radius r and 2r at angular speed.
The acceleration of ball B is
1.
2.
3.
4.
four times as great
twice as great
equal to
one half as great
as the acceleration of ball A.
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The two identical balls are whirled around in a circle as
before. Assume that the balls are moving very fast and that
the two strings are massless.
The tension in string P is
1.
2.
3.
less than
equal to
greater than
the tension in string R.
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Car on circular track with friction
• Motion around circular track, constant speed (for now):
arad = v2/r
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Motion of car on banked
circular track (no friction)
car
N
R
q
a=
W
Speed v
Horizontal forces:
Vertical:
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Motion on loop-the-loop
What is normal force on car at
top and bottom of loop?
Neglect friction; assume
moves with speed vB at
bottom and vT at top
At bottom
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car
At top
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Impulse
• Constant force F12 acting on object 1
due to object 2 for a time Dt yields an
impulse
I12 = F12 Dt
• In general, for a time varying force need to
use this for small Dt and add:
I =  F(t)Dt =
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Impulse for time varying forces
F(t)
* area under curve
equals impulse
t1
Physics 215 – Fall 2014
t2
t
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Impulse  change in momentum
• Consider first constant forces ...
• Constant acceleration equation:
vf = vi + at
mvf - mvi = mat =
• If we call p = mv momentum we see that
Dp =
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Impulse demo
• Cart equipped with force probe collides
with rubber tube
• Measure force vs. time and momentum
vs. time
• Find that integral of force curve is
precisely the change in p!
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Definitions of impulse and momentum
Impulse imparted to object 1 by object 2:
I12 = F12 Dt
Momentum of an object:
p = mv
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Impulse-momentum theorem
Inet = Dp
The net impulse imparted to an object is
equal to its change in its momentum.
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Consider the change in momentum in these three cases:
A. A ball moving with speed v is brought to rest.
B. The same ball is projected from rest so that it moves
with speed v.
C. The same ball moving with speed v is brought to rest
and immediately projected backward with speed v.
In which case(s) does the ball undergo the largest
change in momentum?
1.
2.
3.
4.
Case A.
Case B.
Case C.
Cases A and B.
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Throwing ball at student on cart
(Neglecting vertical forces for short time interval
Physics 215 – Fall 2014
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while student catches ball.)
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Notice
• By Newton’s 3rd law, the force on the
ball is equal and opposite to the force
on the student
• Acts for same time interval  equal and
opposite changes in momentum
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Newton’s 3rd law
and changes in momentum
F1,2 = -F2,1
If all external forces (weight, normal, etc.) cancel:
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Conservation of momentum
• Assuming no net forces act on bodies
there is no net impulse on composite
system
• Therefore, no change in total
momentum D(p1+ p2) = 0
1
Physics 215 – Fall 2014
F
2
system
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Conservation of momentum
(for a system consisting of two objects 1 and 2)
If the net (external) force on a system is zero,
the total momentum of the system is constant.
Whenever two or more objects in an isolated
system interact, the total momentum of the
system remains constant
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Conservation of momentum with carts
• Two carts initially at rest with compressed
spring between them -- track with motion
detectors
pi = 0
pf =
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A cart moving to the right at speed v collides
with an identical stationary cart on a low-friction
track. The two carts stick together after the
collision and move to the right.
What is their speed after colliding?
1.
2.
3.
4.
0.25 v
0.5 v
v
2v
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A student is sitting on a low-friction cart and is
holding a medicine ball. The student then throws
the ball at an angle of 60° (measured from the
horizontal) with a speed of 10 m/s.
The mass of the student (with the car) is 80 kg.
The mass of the ball is 4 kg.
What is the final speed of the student (with car)?
1.
2.
3.
4.
0 m/s
0.25 m/s
0.5 m/s
1 m/s
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Momentum is a vector!
pA,initial + pB,initial = pA,final + pB,final
• Must conserve components of momentum
simultaneously
• In 2 dimensions:
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At the intersection of Texas Avenue and University Drive, a blue, subcompact car
with mass 950 kg traveling east on University collides with a maroon pickup truck
with mass 1900 kg that is traveling north on Texas and ran a red light. The two
vehicles stick together as a result of the collision and, after the collision, the
wreckage is sliding at 16.0 m/s in the direction 24o east of north. Calculate the speed
of each vehicle before the collision. The collision occurs during a heavy rainstorm;
you can ignore friction forces between the vehicles and the wet road.
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Reading assignment
• More impulse and momentum, collisions
• Remainder of Chapter 9 in textbook
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